Arc Length Reparameterization (Ex2)

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How to find the Arc Length of a Curve – Re-parameterization a Curve (Example 2)

Here’s another arc length calculus example problem, but this time we’re asked to “reparameterize” a curve. Before continuing, we recommend you check out our example problem on how to find the regular arc length of a curve without reparameterization. 

What this complicated term “reparameterization” really just means is to just find r(t(s)), which is r in terms of s, and s itself is in terms of t.

It’s somewhat like “function composition” you may have done earlier, where you substitute one function into another. 

Example Problem: Reparameterize the curve r(t) with respect to arc length, starting from t = 0 to a direction of increasing t, where \(r(t) = 2ti + (1 – 3t)j + (5 + 4t)k\)

Arc length is given by (memorize this or ask your teacher/professor if it’s on a cheat sheet or provided during tests/exams):

\[s(t) = L = \int_0^t {\left| {r'(u)} \right|} du\]

Note that the equation above is for r(t), but you’re not technically allowed to have the same variable for r as what is in the integral boundaries (t).

Even though it says r(u), it really is just r(t). They just used another variable because of nomenclature reasons, but it’s just r(t), do not over-complicate it. 

We start by finding the derivative of r(t).

\[r'(t) = 2i – 3j + 4k\]

Then substitute it into the arc length formula. You will always be asked to determine arc length on a “closed interval”, meaning the boundaries on the integral are just the endpoints of the domain you’re provided.

\[s(t) = \int_0^t {\sqrt {{{(2)}^2} + {{( – 3)}^2} + {{(4)}^2}} du} \]

\[s(t) = \sqrt {29} (t – 0)\]

\[s(t) = \sqrt {29} t\]

Rearrange in terms of t, and substitute it into r(t). That’s just what arc length parameterization is.

\[t = \frac{s}{{\sqrt {29} }}\]

And our original r(t) is:

\[r(t) = 2ti + (1 – 3t)j + (5 + 4t)k\]

Substituting in we obtain:

\[r\left( {t(s)} \right) = \frac{{2s}}{{\sqrt {29} }}i + \left( {1 – \frac{{3s}}{{\sqrt {29} }}} \right)j + \left( {5 + \frac{{4s}}{{\sqrt {29} }}} \right)k\]

Done! Sometimes the final substitution gets messy arithmetically.

How to plot parametric curves in MATLAB

Are you curious what the curve in the question looks like? It’s from here onward that your mathematics courses are probably starting to look more and more abstract. BUT, it doesn’t have to be that way!

It’s usually painstaking to plot these by hand because it takes a significant amount of time to do computations to gather the points. But with computational software such as MATLAB, you can easily make a script to sketch the curve very easily.

Use the following MATLAB script. You can use it for other parametric curves as well, just replace the three equations with what you were given, and also replace the domain in the last line of code. We just used from t = 0 to 10 since it said in the direction of increasing t.

syms t
xt = 2*t;
yt = 1-3*t;
zt = 5+4*t;
fplot3(xt,yt,zt,[0 10])
xlabel('x')
ylabel('y')
zlabel('z')

You will obtain the following plot:

how to plot parametric curves in MATLAB

The plot is just a line in 3D because each component in r(t) is just a line. It doesn’t look as curvy as our previous example because we don’t have trigonometric terms in the r(t) equation.

Move to another topic, conservative vector fields, and finding potential and work

Return to previous example problem finding arc length of a curve.

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